Answer :

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k - 1, 5k) are collinear.

Then, area of ΔABC = 0

1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{1} - y_{2} )] = 0

Here,

x_{1} = k + 1,

x_{2} = 3k

x_{3} = 5k – 1

And

y_{1} = 2k

y_{2} = 2k + 3

y_{3} = 5k

[(k + 1)(2k + 3 - 5k) + 3k(5k – 2k) + (5k – 1)(2k + 3)]

[(k + 1)( - 3k + 3) + 3k(3k) + (5k - 1)(2k – 2k – 3)] = 0

[ - 3k^{2} + 3k - 3k + 3 + 9k^{2} - 15k + 3] = 0

(6k^{2} – 15k + 6) = 0 (multiply by 2)

→ 6k^{2} – 15k + 2 = 0

→ 2k^{2} – 5k + 2 = 0

[Divide by 3]

→ 2k^{2} – 4k – k + 2 = 0

→ 2k (k – 2) – 1(k – 2) = 0

→ (k – 2)(2k – 1) = 0

If k – 2 = 0, then k = 2

If 2k – 1 = 0, then k= 1/2

∴ k = 2, 1/2

Hence, the required values of k are 2 and 1/2.

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